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Solution to diff. eqn as a ring

  1. Sep 29, 2008 #1
    Solution to diff. eqn as a ring(killing me - please look at my result!)

    1. The problem statement, all variables and given/known data

    I am presented with the following problem given the direction field

    [tex]f(u,v) = \left( \begin{array}{ccc} -sin(u) & sin(v) \\ cos(u) & cos(v) \end{array} \right)[/tex]

    as [tex]\left( \begin{array}{c}u^{'} & v^{'} \end{array} \right) = f(u,v)[/tex]

    show that

    [tex](\{(u,v) \in \mathbb{R}^2|f(u,v) = 0\} = T_{1} \cup T_{2} )[/tex]


    [tex]T_{1} = \{l \pi, m\pi + \frac{\pi}{2}|l,m \in \mathbb{Z}\}[/tex]
    [tex]T_{2} = \{m \pi, l\pi + \frac{\pi}{2}|l,m \in \mathbb{Z}\}[/tex]

    3. The attempt at a solution

    Don't I treat [tex]T_{1} \cup T_{2} [/tex] as a ring since any solution to the differential equation [tex]\left( \begin{array}{c}u^{'} & v^{'} \end{array} \right) = f(u,v) = 0[/tex] must lie on the Ring [tex]T_{1} \cup T_{2} [/tex]?

    Thus I must show [tex]T_{1} + T_{2} [/tex] and [tex]T_{1} \cdo T_{2} [/tex] lies
    [tex]T_{1} \cup T_{2} [/tex]?

    If yes very well.

    [tex]T_{1} + T_{2} = (\frac{4(l+1)\cdot \pi}{2}, \frac{4(m+1)\cdot \pi}{2}) [/tex]


    [tex]T_{1} \cdot T_{2} = (\frac{(2 \cdot m+1)\cdot \pi}{2}, \frac{(2\cdot l+1)\cdot \pi}{2}) [/tex]

    and since [tex]l,m \in \mathbb{Z}[/tex] then 2(l+1) and 2(m+1) and m(2m+1) and l(2l+1) are in [tex]T_{1} \cup T_{2} [/tex] is the Ring which defines in solution for [tex]\left( \begin{array}{c}u^{'} & v^{'} \end{array} \right) = f(u,v)[/tex] which implies that in l,m on the Ring mentioned above will give a solution for the differential equation.

    How is that??

    Sincerely Yours

  2. jcsd
  3. Oct 1, 2008 #2


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    Science Advisor
    Homework Helper

    Hi Alphaboy! Thanks for the PM :smile:
    I'm sorry, I'm not following this. :redface:

    T1 and T2, according to your definitions, are the same.

    And how can f(u,v) = 0? If the top line is 0, then the bottom line isn't.

    And what is the differential equation?

    And what is a direction field? :confused:
  4. Oct 1, 2008 #3
    Hi Tiny-tim,

    you are one of the smartest guys here :D

    Anyway if look at my teachers definition of T1 and T2 its their inclusion which gives f(u,v) = 0 for any integer l and m, if and only if l and m are in Z.

    and what I am trying to do it to insert T1 + T2 and T1*T2 into f(u,v) and as mentioned if l and m are in Z, then I get f(u,v) = 0.

    and it was "vectorfield" and (u', v') is the corresponding differential equation.

    I hope this have cleared it up now :)

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